Question

(a) State Kirchhoff's rules for an electric network Using Kirchhoff's rules , obtain the balance condition in terms of the resistance of four arms of Wheat stone bridge
(b) In the meter bridge set up, shown in the figure, the null point 'D" is obtained at a distance of 40 cm from end A of the meter bridge wire. if a resistance of $\Omega$ is connected in series with $R_1$ null point is obtained at AD=60 cm . Calculate the values of $R_1$ and $R_2$.

Answer 1

A. In complicated networks to find current flowing through different branches we apply Kirchhoff's law.

First law: The algebraic sum of currents meeting at a point is zero. i.e.Σi = 0.

In the given circuit, Considering outward direction of current as positive and inward as negative then at the point F

According to Kirchhoff’s first law, at junction F

$I_1+I_2=I_3$

Second law: In a closed mesh of electrical conductors the algebraic sum of product of resistance and the respective current in the different branches is equal to the total e.m.f applied in the closed mesh.

$\sum IR=\sum E$

Using Kirchhoff’s law we now find the condition for null deflection in a Wheatstone bridge

Four resistance P,Q,R and S are joined end to end to form a closed circuit. These close networks of conductors form a closed circuit. This close network of conductors is known as Wheatstone bridge. Between any pair of opposite junctions say A & C a battery is connected and between the other pair of opposite junctions a galvanometer is connected. We assigned the current flowing in the different branches from the logical consideration.

Given : G = The resistance of the galvanometer Let i1,i2,i3,i4 & ig be the current flowing through the branches of resistance P, R, Q, S and G respectively.

Applying Kirchhoff's first law :

(1) At the point C:

$\sum i=i-i_1{-i}_2=0\Rightarrow i_2=i-i_1\left(1\right)$

$i_3=i_1-i_g\left(2\right)$

$i4=i-i_1+i_g\left(3\right)$

Applying Kirchhoff's second law :

In closed mesh CBDC:

$i_{1}P+i_{g}G-i_{2}R=0$

$(P+Q)i_1+i_{g}G-iR=0\left(4\right)$

By applying same in

(ii) In closed mesh BADB

$Q{i}_3-i_{g}G-i_{4}S=0\left(5\right)$

By putting value of (2) in (5) , and eliminating $i_1$ we get

$i_g=\frac{i(RQ-SP)}{G(Q+S)+(Q+G+S)(P+R)}$ (6)

From equation (6) we can find the current flowing through the galvanometer. For null deflection ig=0 from equation (6) ig=0 only if (RQ‐PS)=0 or P/Q=R/S is the condition for null deflection.

B).

$\frac{R_1}{R_2}=\frac{AD}{DC}=\frac{40}{60}\left(1\right)Also\frac{R_1+10}{R_2}=\frac{60}{40}\left(2\right)Dividing\left(2\right)by\left(1\right)\frac{R_1+10}{R_1}=\frac{60}{40}X\frac{60}{40}=\frac{9}{4}$

R1= 8 Ω

R2 = 12Ω